# RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS

## RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS

Other Exercises

Question 1.
If the ratio of mode and median of a certain data is 6 : 5, then find the ratio of its mean and median.
Solution:
We know that
Mode = 3 median – 2 mean…(i)
and $$\frac { mode }{ median }$$ = $$\frac { 6 }{ 5 }$$
Mode = $$\frac { 6 }{ 5 }$$median
∴From (i), $$\frac { 6 }{ 5 }$$ median = 3 median – 2 mean
=> 2 mean = 3 median – $$\frac { 6 }{ 5 }$$median
2 mean = $$\frac { 15-6 }{ 5 }$$median = $$\frac { 9 }{ 5 }$$median
$$\frac { mean }{ median }$$ = $$\frac { 9 }{ 5X2 }$$ = $$\frac { 9 }{ 10 }$$
∴Ratio = 9:10

Question 2.
If the mean of x + 2, 2x + 3, 3x + 4, 4x + 5 is x + 2, find x.
Solution:
Mean of x + 2, 2x + 3, 3x + 4, 4x + 5 = x + 2
=> $$\frac { x + 2+2x + 3+3x + 4+4x + 5 }{ 4 }$$ = x + 2
=> 10x + 14 = 4x + 8
=> 10x – 4x = 8 – 14
=> 6x= – 6
∴ x = – 1

Question 3.
If the median of scores ,$$\frac { x }{ 2 }$$, $$\frac { x }{ 3 }$$, $$\frac { x }{ 4 }$$, $$\frac { x }{ 5}$$ and $$\frac { x }{ 6 }$$ (where x > 0) is 6, then find the value $$\frac { x }{ 6 }$$
Solution:
$$\frac { x }{ 2 }$$, $$\frac { x }{ 3 }$$, $$\frac { x }{ 4 }$$, $$\frac { x }{ 5}$$, $$\frac { x }{ 6 }$$
Here n = 5
Median = $$\frac { n+1 }{ 2 }$$ th term = $$\frac { 5+1 }{ 2 }$$ th
$$\frac { 6 }{ 2 }$$ = 3rd term = $$\frac { x }{ 4 }$$
$$\frac { x }{ 4 }$$ = 6 => x = 24
$$\frac { x }{ 6 }$$ = $$\frac { 24 }{ 6 }$$ = 4
∴Hence = $$\frac { x }{ 6 }$$ = 4

Question 4.
If the mean of 2, 4, 6, 8, x, y is 5, then find the value of x + y.
Solution:
Mean of 2, 4, 6, 8, x, y is 5
$$\frac { 2+4+6+8+x+y }{ 6 }$$ = 5
$$\frac { 20+x+y }{ 6 }$$ = 5
=> 20 + (x +y) = 30
=> x + y = 30 – 20 = 10
∴x + y = 10

Question 5.
If the mode of scores 3, 4, 3, 5, 4, 6, 6, x is 4, find the value of x.
Solution:
Mode of 3, 4, 3, 5, 4, 6, 6, x is 4
∴ 4 comes in maximum times
But here ,
3 2
4 2
5 1
6 2
3, 4 and 6 are equal in number
∴ x must be 4 so that it becomes in maximum times

Question 6.
If the median of 33, 28, 20. 25, 34, x is 29. find the maximum possible value of x.
Solution:
Median of 33, 28, 20, 25, 34, x is 29
Now arranging in ascending order 20, 25, 28, x, 33, 34
Here n = 6
Median = $$\frac { 1 }{ 2 } \left[ \frac { 6 }{ 2 } th\quad term+\left( \frac { 6 }{ 2 } +1 \right) th\quad term \right]$$
29 = $$\frac { 1 }{ 2 }$$ [3rd term + 4th term]
29 = $$\frac { 1 }{ 2 }$$ [28+x]
58 = 28 + x
=> x = 58 – 28 = 30
∴Possible value of x = 30

Question 7.
If the median of the scores 1, 2, x, 4, 5 (where 1 <2 <x <4 <5) is 3, then find the mean of the scores.
Solution:
Scores are 1, 2, x, 4, 5 and median 3
Here n = 5 which is odd
Median = $$\frac { n+1 }{ 2 }$$ th term = $$\frac { 5+1 }{ 2 }$$ = $$\frac { 6 }{ 2 }$$ th
=> 3 = 3rd term = x
=> 3 = x
∴ x = 3
Mean of the score = $$\frac { 1+2+3+4+5 }{ 5 }$$ = 3

Question 8.
If the ratio of mean and median of a certain data is 2 : 3, then find the ratio of its mode and mean.
Solution:
We know that mode = 3 median – 2 mean

$$\frac { mode }{ mean }$$ = $$\frac { 5 }{ 2 }$$
Ratio in mode and mean = 5 : 2

Question 9.
The arithmetic mean and mode of a data are 24 and 12 respectively, then find the median of the data.
Solution:
Mean = 24
Mode = 12
We know that mode = 3 median – 2 mean
12 = 3 median – 2 x 24
12 = 3 median – 48
3 median 12 + 48 = 60
Median = $$\frac { 60 }{ 3 }$$ = 20

Question 10.
If the difference of mode and median of a data is 24, then find the difference of median and mean.
Solution:
Mode – Median = 24
Mode = 24 + median
But mode = 3 median – 2 mean
3 median – 2 mean = 24 + median
3 median – median – 2 mean = 24
=> 2 median – 2 mean = 24
=> Median – Mean = 12 (Dividing by 2)

Hope given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS are helpful to complete your math homework.

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